Practice Paper – TERM II (2021 – 2022)
Class – IX
Mathematics
Time: 2 hours Maximum Marks: 40
General Instructions:
1. The question paper consists of 14 questions divided into 3 sections A, B, C.
2. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
3. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
4. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study based questions.
SECTION – A
1. Two coins are tossed simultaneously 500 times, following are the outcomes No head= 100 times One head = 200 times Two heads = 200 times If the two coins are simultaneously tossed again, compute the probability of obtaining:
i. One Head
Total number of outcomes n(S) = 500
Let E1 and E2 be the events of one head and two heads respectively
Favourable outcomes n(E1) = 200
Then, P(one head) = n(E1)/ n(S)
=200/500 = 2/5
ii. Two Heads
Favourable outcomes n(E2) = 200
Then, P(two head) = n(E2)/ n(S)
=200/500 = 2/5
2. Teachers and students are selected at random to make two teams of 30 members each on sports day to participate in the event of “tug of war”. The number of volunteers are as follows:
Find the probability that the person chosen at random
(i) is a male
Total number of volunteers i.e., n(S)
= 12 + 18 + 20 + 10 = 60
Total number of males i.e., n(E1)
12 + 20 = 32
P(volunteer is male) = n(E1)/ n(S) = 32/60 = 8/15
(ii) is a female student
P(volunteer is female student) = n(E2)/ n(S) = 10/60 = 1/6
OR
In a one-day cricket Match, Sachin played 40 balls and hit 12 sixes and Saurav played 30 balls and hit 9 fours. Find the probability that Sachin will hit a six in the next ball and also find the probability that Saurav will not hit a four in the next ball.
Total number of balls faced by Sachin = 40
Number of balls on which he hit a six = 12
Let E1 be the event of hitting a six.
⸫ Number of outcomes = 12
⸫ P(E1) = 12/40=3/10 = 0.3
Now. Total number of balls faced by Saurav = 30
Let E2 be the event of Saurav did not hit a four
Number of outcomes = 30 – 9 = 21
P(E2) = 21/30 = 7/10 = 0.7
3. A rectangular piece of paper is 22 cm long and 10 cm wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder.
According to the question,
2πr = 22 (Circumference of cylinder)
2 x 22/7 x r = 22
r = 7/2
Volume of the cylinder = πr2h
= 22/7 x 7/2 x 7/2 x 10
= 385 cm3
4. The angles of a quadrilateral are 4x°, 7x°, 15x° and 10x°. Find the smallest and largest angles of the quadrilateral.
Sum of the angles of a quadrilateral is 3600
⸫ 4x + 7x + 15x + 10x = 360
Or 36x = 360
Or x = 360/36
X = 10
Smallest angle = 4x = 4 x 10 = 40
Largest angle = 15x = 15 x 10 = 150
5. If y = 2 and y = 0 are the zeroes of the polynomial f(y) = 2y3 – 5y2 +ay + b, find the value of a and b.
6. A chord of length 10 cm is at a distance of 12 cm from the centre of a circle. Find the radius of the circle.
OR
In the given figure, find the value of x.
SECTION – B
7. If f(x) = 5x2 – 4x+ 5, find f(1) + f(– 1) +f(0).
8. Construct ∠POY = 30°, using compass and ruler
9. If ab + bc + ca = 0, then find the value of
10. Find the radius of the base of a right circular cylinder whose curved surface area is 2/3 of the sum of the surface areas of two circular faces. The height of the cylinder is given to be 15 cm.
OR
The radius and slant height of a cone are in the ratio 4: 7. If its curved surface area is 792 cm2, find its radius.
Let the radius of cone be r = 4x and slant height l = 7x
CSA of a cone = 792 cm2
πrl = 792
or 22/7 x 4x x 7x = 792
x2 = (792 x 7)/(22 x 4 x 7) = 9
x = 3
radius = 4 x 3 = 12 cm
SECTION – C
11. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
12. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being4.4 cm. Find its: (i) inner curved surface area
Inner radius (r) = 2 cm
Outer radius (R) = 2.2 cm
Height (h) = 77 cm
CSA (inner) = 2πrh
=2 x 22/7 x 2 x 77
= 968 cm2
(ii) outer curved surface area
CSA (outer) = 2πRh
= 2 x 22/7 x 2.2 x 77
= 1064.8 cm2
(iii) total surface area. (use π= 22/7)
Area of top = π(R + r)(R – r)
= 22/7 x 4.2 x 0.2
= 2.64 cm2
= Area of the bottom
⸫T.S.A = Inner (C.S.A) + Outer (C.S.A) + Area of top + Area of bottom
= 968 + 1064.8 + 2 + 2.64
= 2038.08 cm2
OR
The frame of a lampshade is cylindrical in shape. It has base diameter 28 cm and height 17 cm. It is to be covered with a decorative cloth. A margin of 2 cm is to be given for folding it over top and bottom of the frame. If 1/12 of cloth is wasted in cutting and pasting, find how much cloth is required to be purchased for covering the frame
Base diameter = 28 m
Base radius = 14 cm
Height of cloth required = 17 + 2 + 2 = 21 cm
Area of cloth required = curved surface area of cylinder of radius 14 cm and height 21 cm
= 2πrh
= 2 x 22/7 x 14 x 21
1848 cm2
Let A sq. cm of cloth be purchased
So, wastage of cloth for cutting and pasting = A/12 cm2
Area of cloth actually used = A – A/12 = 11A/12 cm2
Area of cloth actually used = Area of cloth required
Or, 11A/12 = 1848
Or A = (1848 x 12)/11 = 2016 cm2
Case Study 1
13. Read the following text and answer the following questions on the basis of the same:
Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls
In a school, a group of (x + y) teachers, (x2 + y2) girls and (x3 + y3) boys organised a campaign on Beti Bachao, Beti Padhao.
(i) If in the group, there are 10 teachers and 58 girls, then what is the number of boys?
(ii) If x –y = 23, then find x2 –y2.
Case Study 2
14. Read the following text and answer the questions given below:
National Association for the Blind (NAB) aimed to empower and well-inform visually challenged population of our country, thus enabling them to lead a life of dignity and
productivity. Ravi donated Rs. (x3 + 1/x3) to NAB. When his cousin asks to tell the amount donated by him, he just gave, the hint x + 1/x = 10
(i) Find the amount donated by Ravi.
(ii) Find the amount donated by Ravi if x + 1/x = 7
